Question

The least integer $x$ for which the inequality $\frac{{(x-3)}^2}{x^2+8x-22}<0$ is satisfied, is

• A. $-8$
• B. $-9$
• C. $-10$
• D. $-11$

Answer 1

The correct option is A $-8$

$\frac{(x-3{)}^2}{x^2+8x-22}<0$

As $(x-3{)}^2=0$ at $x=3$

so, $x\ne 3$

Except $3,(x-3{)}^2>0\forall x\in R$

$\Rightarrow x^2+8x-22<0$

$\Rightarrow x^2+8x+16-16-22<0$

$\Rightarrow (x+4{)}^2-38<0$

$\Rightarrow |x+4|<38$

$\Rightarrow (x+4)\in (-\sqrt{38},\sqrt{38})$

$\Rightarrow x\in (-\sqrt{38}-4,\sqrt{38}-4)$

Least integer in this interval is -8.